Question

An aeroplane is flying parallel to the Earth's surface at a speed of 175m/sec and at a height of 600m. The angle of elevation of the aeroplane fron a point on the Earth's surface is 37° at a given point. After what period of time does the angle of elevation increases to 53°?

Answer 1

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Step-by-step explanation:

Answer 2

Answer:

Let C is the initial and D is the final position of the aeroplane.

Let the time taken by the aeroplane be t.

∴ CD = 175 t ... (Distance = speed × Time)

Let AB be x

∴ AE = x + 175 t

In the right ∆ ABC

tan 53° = BC / AB

✈ 1.3270 = 600 / x

x = 600 / 1.237 -----> EQUATION 1

In the right ∆AED, tan 37° = DE / AE

0.753 = 600 / x + 175 t

x + 175 t = 600 / 0.7536

x = 600 / 0.7536 - 175 t -------> EQUATION 2

From 1 and 2 we get,

600 / 1.327 = 600 / 0.7536 - 175 t

175 t = 600 / 0.7536 - 600 / 1.327

175 t = 796.18 - 452.15 = 344.03

∴ t = 344.03 / 175 = 1.97 seconds.

∴ Time Taken is 1.97 seconds.

Step-by-step explanation:

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